Answer
15.7 M
Work Step by Step
Mass fraction of $HNO_{3}$= weight percent$ \div100$
$=70/100= 0.70=\frac{mass\,of\,HNO_{3}}{mass\,of\, solution}$
Let the volume of the solution be 1 L=1000 mL.
Then, mass of the solution=$ 1.41\,g/mL\times1000\,mL=1410\,g$
Mass of $HNO_{3}$ = mass of solution$\times$mass fraction of nitric acid
$=1410\,g\times0.70= 987 g$
Moles of nitric acid=$ \frac{mass\,of \, nitric\, acid}{molar\,mass}=\frac{987\,g}{63.0\,g/mol}=15.7\,mol$
Molarity=$ \frac{moles\,of\, nitric\, acid}{volume\, of\, solution\,in\,L}=\frac{15.7\,mol}{1\,L}= 15.7 M$
